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date: 2022-02-23 18:45
modification date: Friday 18th March 2022 15:19:48
title: "Hopf invariant one"
aliases: [Hopf invariant]

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# Hopf invariant

Let $f: S^{2n-1} \to S^n$, then $\cone(f)$ has a CW structure $e^0 \union e^n \union e^{2n}$. 
Let $H^n(\cone(f)) = \alpha\ZZ$ and $H^{2n}(\cone(f)) = \beta \ZZ$ where $\alpha = [e^n]\dual$ and $\beta = [e^{2n}]\dual$.
Then $\alpha^2 \in H^{2n}(\cone(f))$ so $\alpha^2 = n\beta$ for some $n\in \ZZ$; this $h(f) \da n$ is the Hopf invariant of $f$.

Alternatively, take $f$ smooth and $\vol \in \Omega^n(S^n)$ a volume form.
Since $\int_{S^n} \vol = 1$, the pullback $f^*\vol$ is exact since $H^n(S^{2n-1}) = 0$, so $f^*\vol = d\alpha$ for some $\alpha \in \Omega^{n-1}(S^{2n-1})$.
Then
$$
h(f) = \int_{S^{2n-1}} \alpha\wedge \dalpha
$$


# Hopf invariant one

![](attachments/Pasted%20image%2020220403214059.png)

# Motivation

Hopf famously discovered an essential (i.e. non-nullhomotopic) map $S^3\to S^2$, which you can think of as the being defined via $S^3\to \CP^1$ and then choosing an identification of $\CP^1$ with the sphere. The [mapping cone](mapping cone.md) of this map is a CW complex and asking whether or not the map $S^3\to S^2$ is non-nullhomotopic is equivalent to asking whether or not this mapping cone is homotopy equivalent to a wedge $S^4\vee S^2$. 

We can see that it is not since the mapping cone is actually just $\CP^2$ and the cup square of the generator in $H^2$ is the generator in $H^4$. More generally, given a map $S^{2n−1}\to S^n$ we can form the 2-cell complex given by gluing $\DD^{2n}$ along this map and ask if the cup square of the generator in degree $H^n$ squares to the generator in degree $2n$. 

If it does we say this map has **Hopf invariant 1**, otherwise it has Hopf invariant 0. It's natural to ask **how many maps of Hopf invariant one can we build**? The answer is: **not many**. They only exist when $n=1,2,4,8$. While nowadays we usually learn the proof of this fact using [K-theory](Unsorted/K-theory.md), which is very short.